Least squares fitting with elliptic paraboloids

Transcription

1 MATHEMATICAL COMMUNICATIONS 409 Math. Coun. 18(013), Least squares fitting with elliptic paraboloids Heluth Späth 1, 1 Departent of Matheatics, University of Oldenburg, Postfach 503, D Oldenburg, Gerany Received April 6, 013; accepted June 0, 013 Abstract. In [4], we discussed the proble of fitting soe rotated paraboloid to given easured data in 3-space. A nuerical ethod was developed and nuerical exaples were given. Recently that ethod was used to fit soe parabolic antenna filter [1]. In this connection the question arises whether this ethod could be extended for elliptic paraboloids needed for soe edical application []. We show how this could be done and recoend a nuerical ethod ore difficult but siilar to that one in [4]. AMS subject classifications: 65D10, 65C60 Key words: least squares fitting, elliptic paraboloids, nuerical ethod 1. The proble An elliptic paraboloid with the z-axis as the axis and the origin as the vertex is given by ( ) x z = d A + y B, (1) where A and B are the half axes of an ellipse and d is varying. For A = B = 1 we would have a rotated paraboloid [4]. Instead of (1), we will better use the canonical paraetric for x = Av cos u y = Bv sin u () z = dv, where u [0, π), < v <, If (a, b, c) T is soe shifting of the origin and cos β 0 sin β Q(β) = (3) sin β 0 cos β P (γ) = cos γ sin γ (4) 0 sin γ cos γ Corresponding author. Eail address: spaeth@atheatik.uni-oldenburg.de (H. Späth) c 013 Departent of Matheatics, University of Osijek

2 410 H. Späth are rotations in the x z plane and in the y z plane, then our extended odel () reads x a + Av cos u y = P (γ)q(β) b + Bv sin u. (5) z c + dv Reark: A third rotation could be included in the x z plane [], but this is oitted due to a larger extent. If now data points (x i,, ) T (i = 1,..., ) are given to which odel (5) should be fitted in the least square sense, then 1 x i P (γ)q(β) a + Av i cos u i b + Bv i sin u i c + dvi has to be iniized with respect to the 8 + unknowns a, A, b, B, c, d, β, γ, u, v, where u = (u 1,..., u ) T, v = (v 1,..., v ) T. (7) It turns out to be soewhat ore convenient transforing the given data (x i,, ) T by x i x i x i x i = P (γ) T, ỹ i = Q(β) T. (8) Note that P (γ) 1 = P (γ) T and Q(β) 1 = Q(β) T. Now the iniization of (6) is equivalent to the iniization of the following objective S(a,A, B, b, c, d, β, γ, u, v) = 1 x i a + Av i cos u i ỹ i b + Bv i sin u i c + dvi = 1 ( x i a Av i cos u i ) + (ỹ i b Bv i sin u i ) + ( c dvi ) (9) = 1 S i (a, A, b, B, c, d, β, γ, u i, v i ).. Necessary conditions for iniu All 8 + partial derivatives of S with respect to the 8 + unknowns ust becoe zero. This is a highly nonlinear syste of equations that can be solved neither explicitly nor easily nuerically (e. g. by NEWTON s ethod). Instead we will use a successive iniization ethod which is generally described in [4] and also realized for rotated paraboloids. But at first we explicitly need all partial derivatives of S. We will list the by the order of variables given in (7). (6)

3 Least squares fitting with elliptic paraboloids 411 We have a = 0 A = 0 b = 0 B = 0 c = 0 d = 0 a + Av i cos u i = av i cos u i + Av i cos u i = b + Bv i sin u i = bv i sin u i + Bv i sin u i = c + dv i = cv i + dv 4 i = x i, (10) v i cos u i, (11) ỹ i, (1) v i sin u i (13) (14) vi. (15) Further, = 0 + ( x i a Av i cos u i ) + Here corresponding to (8) we have (ỹ i b Av i sin u i ) ( c dv i ) = 0. (16) x i ỹ i = Q(β) T P (γ) T x i (17) and also z i = Q(β)T x i. (18) Inserting (17) and (18) into (16) we receive = 0 H sin β G cos β = 0, (19)

4 41 H. Späth where H = G = x i (a + Av i cos u i ) + (c + dvi ) (a + Av i cos u i ) + x i (c + dvi ). For a iniu we ust have In this case S = H cos β + G sin β > 0. β = atan ( ) G, (0) H else this value has to be replaced by β + π. Using (8) the variable β also depends on γ. Siilarly to (16)-(0) we have = 0 + ( x i a Av i cos u i ) + (ỹ i b Bv i sin u i ) ( c dv i ) = 0. (1) Using (8) we have z i T P (γ)t = Q(β) x i. () Putting (17) and () into (1) after lengthy calculations we get where U = V = = 0 U cos γ + V sin γ = 0, (3) sin β (a + Av i cos u i ) (b + Bv i sin u i ) + cos β (c + dvi ), sin β (a + Av i cos u i ) + (b + Bv i sin u i ) + cos β (c + dvi ).

5 For a iniu we ust have In this case Least squares fitting with elliptic paraboloids 413 S = U sin γ + V cos γ > 0. γ = ( ) U V else this value has to be replaced by γ + π. Finally, = = 0 (i = 1,..., ) v i v i ust be exained. First we consider v i = i v i = 0 A cos u i ( x i a Av i cos u i ) + B sin u i (ỹ i b Bv i sin u i ) (4) + dv i ( c dv i ) = 0 (5) d v 3 i (A cos u i + B sin u i d( c))v i [A cos u i ( x i a) + B sin u i (ỹ i b)] = 0. These are polynoial equations of degree three, each with one or three real solutions. If there are three real roots, then the one with the sallest value for S i (i = 1,..., ) ust be selected. Analogously, we have to look for Putting for (6) we get Substituting u i = i u i = 0 Av i sin u i ( x i a Av i cos u i ) Bv i cos u i (ỹ i b Bv i sin u i ) = 0 (6) (B A )v i sin u i cos u i + Av i sin u i ( x i a) Bv i cos u i (ỹ i b) = 0. u = u i, w = (B A )v i, p = Av i ( x i a), q = Bv i (ỹ i b) w sin u cos u + p sin u q cos u = 0. (7) r = tgu for (7) we receive again a polynoial equation but now of degree four. It can be shown that it has at least one real root [3]. This eans that there exist two or four real solutions. Again for each i = 1..., this one with the sallest value of S i has to be chosen. Note that polynoial equations of degree three and four can be solved exactly, i.e. without soe iterative nuerical ethod.

6 414 H. Späth 3. Nuerical algoriths There are any possible sequences of the following series of necessary conditions (10) through (15), (18,3), and (5,6) for i = 1,..., to be always fulfilled during the iniization process. We will describe two possible sequences. All values of unknows newly calculated in previous steps ought to be used within Steps 1 4. Method I Step 0: Let starting values for β, γ, u and v be given (e.g. u i = πi, v i equidistant within the interval [in z k, ax z k ] for k = 1..., ). k k Step 1: Calculate (a, A), (b, B), and (c, d) using in each case the corresponding two linear equations (10,11), (1,13), and (14,15). Those will always have a unique solution corresponding to an absolute iniu. Step : Calculate β = β(a, A, c, d, γ) by (19) and also γ = γ(a, A, b, B, c, d, β) by (3) as described. Step 3: Calculate all real zeroes (1 or 3) for v i = v i (u i ) via (5,6) and also those ( or 4) for u i = u i (v i ) (i = 1,..., ) and select in each case the one with the sallest value of S i. Step 4: Calculate S. If it has decreased again, then go back to Step 1, else STOP. Check for soe iniu of S, i.e. if at least all necessary conditions are fulfilled. Method II Step 0: Let a, A, b, B, c, d, β, γ and u or v be given as starting values. Step 1: Calculate u = u(v) or v = v(u) for given v or u in Step 1 siilarly to Step 3 of Method I. Select the zeroes such that S i is inial. Step : Calculate β = β(γ) and γ = γ(β) by (19) and (3) as in Step of Method I. Step 3: Calculate (a, A), (b, B), and (c, d) as in Step 1 of Method I. Step 4: As decribed in Method I. There ay be other sequences for the variables resulting in other ethods, too. But it does not see possible to select soe best sequence in the sense that its success does not depend on the given data (x i,, ) T (i = 1,..., ) and with best convergence properties. There is also no convergence proof for any ethod. But as was indicated by nuerical experients for A = B = 1 [4], Method II perfectly worked and so we recoend it here, too. References [1] Pietro Cerveri, private counication, January 011. [] Guiseppe Orlando, private counication, May 011.

7 Least squares fitting with elliptic paraboloids 415 [3] H. Späth, Least squares fitting of ellipses and hyperbolas, Coput. Statist. 1(1997), [4] H. Späth, Least squares fitting with rotated paraboloids, Math. Coun. 6(001),